Task 1.
The ratio of the intensive use of equipment, capital productivity and capital ratio.

The cost of the equipment shop at the beginning of the year and 17.3 million rubles as of March 1, commissioned equipment worth 440, 9 thousand rubles as of 1 July was eliminated equipment amounting to 30.4 thousand rubles.. output – 800 t-1 t – 30 thousand rubles.. Normative production capacity of 1 thousand tons

Question.
To determine the coefficient of intensive use of the equipment, the return on assets and capital ratio.

Task 2.
The shift factors, loading equipment, extensive, intensive and integrated use of the equipment.

In the shop of the plant installed machines 210. The two-shift mode, shift duration – 8 hours Annual output of 190 000 products, production capacity of the plant is 210 000 products. In the first shift work in all machines, the second is 60% of the machinery, working days in a year – 263, the time of the actual operation of one machine per year is 4000 hours.

Question.
To determine the shift factors, loading equipment, extensive, intensive and integrated use of the equipment.

Task 3.
Determination of the production capacity of the plant.

It is known that in the main production workshop:
– number of machines leading the production at the beginning of the year – 70%;
Nov 1 entered – 25 units;
– two-shift mode;
– duration – 8 hours;
– regulated downtime for equipment repair amount to 7% of the operating Fund operating time of the equipment;
performance 1 machine – 4 parts per hour;
– may 1, left 3 units of equipment;
– working days in the year 260.

Question.
To determine the production capacity of the basic production facility under specified conditions.

Task 4.
The linear method and the withdrawal method value by the sum of years of useful life to determine the annual depreciation.

The initial cost of the machine is 80 000 rubles, the service life is 12 years.

Question.
To determine the annual value of depreciation straight line method and the withdrawal method value by the sum of years of useful life.

Task 5.
The average annual value of fixed assets, cost of funds at the year end, the coefficients of the input and Disposals.

It is known that:
– the value of assets at beginning of year – 9100 thousand rubles;
– March 1, entered 3200 thousand rubles;
– eliminated due to the deterioration of October 1, 4500 thousand roubles, and on December 1 to 700 thousand rubles

Question.
Determine the average annual value of fixed assets, cost of funds at the year end, the coefficients of the input and Disposals for the following data.

In response to successively bring the numbers located at the first position after the decimal point, on the eleventh, twenty-first, thirty-first, forty-first and the last digit is not equal to zero (the numbering of the positions in the decimal system).

8) a Computer network connects computers in different places. Not all computers are connected directly, but each computer can send messages to any other through zero or more intermediate computers. Sending messages from one computer to another, related directly takes one second. The computer cannot send messages to multiple computers simultaneously.

First and second primary computer can send a message to one of its neighbors. Second second 2 computers (home and his neighbor) can send messages to their neighbors and so on, to Determine the minimum time required for transmission of messages from the specified computer to all other computers on the network.

Important skill, without which it is impossible to solve the problem is to determine which section it belongs: mono-, di – or polygyridae crossing; inheritance, sex-linked, or inheritable characteristics when interacting genes… This allows students to choose necessary for solving the problem of laws, regulations, rules, relationships. To this end, you can give the text a task and offer to determine which section it belongs. Students should remember that inherited genes, not the symptoms.

Exercise 1. One of the breeds of chickens differs short legs (these chickens do not break the gardens). This characteristic is dominant. Managing gene causes both a shortening of the beak. However, homozygous chickens beak so small that they are unable to pierce the egg shell and die, not valuewise of eggs. Incubator farms, diluent only short-legged hens received 3000 chickens. How many of them short-legged?

Exercise 2. In medicine is significant difference between the four groups of human blood. Blood group is a hereditary characteristic that is dependent on a single gene. This gene has not two, but three alleles, denoted by a, b, 0. Individuals with genotype 00 have first blood, with genotypes AA or A0 – second, BB or B0 – third, AB – fourth (we can say that alleles a and b dominate allele 0, then as each other, they do not suppress). What blood group may have children if their mothers were the second group of blood, and his father first?

Answer: both tasks on monohybrid crossing, since we are talking about a single gene. (Key words highlighted in the text of the task.)
The types of tasks

All genetic problem, whatever topics they touch (mono – or polygyridae crossing, autosomal or sex-linked inheritance, inheritance mono – or polygenic traits) can be reduced to three types: 1) design; 2) identification of the genotype; 3) to identify the nature of inheritance of the trait.

In the condition of the estimated task must contain:
– about the nature of inheritance (dominant or recessive, autosomal or sex-linked and other);
directly or indirectly (through phenotype) should indicate the genotypes of the parental generation.
The question of the settlement of the problem concerns the prediction of the genetic and phenotypic characteristics of the offspring. For example, the tasks of settlement type.

Task 2. In humans, the gene for polydactyly (Megaplast) dominates over the normal structure of the brush. My wife brush normal, husband heterozygote gene of polydactyly. Determine the probability of birth in this family megapage child.

The solution to this problem begins with the recording of her conditions and designations of genes. Then determined by (presumably) the genotypes of the parents. The genotype of the man known genotype wife is easy to install on phenotype – she is a carrier of the recessive trait, then homozygous for the corresponding gene. The next stage is writing the values of gametes. You should pay attention to the fact that homozygous organism produces one type of gametes, so often found the writing in this case two identical gametes does not make sense. Heterozygous organism produces two types of gametes. The connection of gametes randomly, so the appearance of two types of zygotes is equally probable: 1:1.

Please pay your attention to the inadmissibility to answer in this form: “One child will be born normal and one megapolis” or worse: “the First child will be megapolis, and the second normal”. How many and which children will have spouses, cannot, therefore, it is necessary to operate the concept of probability.
In the problem on determination of the genotype must contain following information:
– about the nature of inheritance of the trait;
– about the phenotypes of the parents;
– about the genotypes of the offspring (directly or indirectly).
The question is this task requires the characteristics of the genotype of one or both parents.

Task 3. In mink brown fur is dominant over blue. Crossed brown female with male blue color. Among the offspring of two puppies brown and one blue. Purebred whether female?

Recordable condition task entering denote genes. Solution start with the design of the crossing. The female has a dominant character. It can be either Homo- (AA) and heterozygous (AA). The uncertainty of the indicated genotype A. The male with the recessive characteristic homozygotes the corresponding gene – AA. Descendants with brown fur inherits this gene from the mother and from the father – gene blue colouring, consequently, their genotypes are heterozygous. The genotype brown puppies to establish the genotype of the mother is impossible. Blue puppy from each parent received a gene blue color. Therefore, the mother heterozygote (nacistoidna).

In terms of tasks to establish the nature of the inheritance:
– offered only phenotypes successive generations (i.e., the phenotypes of parents and phenotypes of offspring);
– contains quantitative characteristics of the offspring.
In this question we will need to establish the nature of inheritance of the trait.

When solving this task, the logic of reasoning could be next. The splitting in the progeny indicates heterozygosity of the parents. A ratio close to 1 : 2 : 1 speaks of heterozygosity for a single pair of genes. According to the obtained fractions (1/4, white, 1/2 motley, 1/4 black), black and white chickens, homozygote and motley are heterozygous.
The designation of genes and genotypes, followed by mapping of the crossing shows that the conclusion corresponds to the crossing.

Answer: colour in chickens is defined by a pair predominantly genes, each of which causes a white or black color, and together they control the development of variegated plumage.
Using the lessons illustrated tasks

Tasks genetics can be divided into text and illustrated. The advantage of the illustrated tasks in front of the text is obvious. It is based on the fact that the visual perception of the image activates the attention and interest of students, contributes to a better understanding of the problem and study patterns.

Task 5.

1. What color fur in rabbits dominates?
2. What are the genotypes of the parents and hybrids of the first generation on the basis of wool dyeing?
3. What genetic regularities appear in this hybridization?

The responses.

1. Dominated by the dark coloration of the coat.
2. P: AA x AA; F1 : Aa.
3. We see manifestations of the rules of dominance characteristics and uniformity of the first generation.

Drawings can be sketchy.

Task 6.

1. What form of fruit of tomato (spherical or pear-shaped) dominates?
2. What are the genotypes of the parents and hybrids 1 and 2 generation?
3. What genetic regularities discovered by Mendel, appear at this hybridization?

The responses.

1. Dominated by the spherical form of the fetus.
2. P: AA x AA; F1 : AA; F2 : 25% AA, 50% AA, 25% AA.
3. The laws of uniformity of hybrids of the first generation (the first law of Mendel) and the law of splitting (the second law of Mendel).

Task 7.

1. What are the genotypes of the parents and F1 hybrids, if the red color and round shape of the fruit of the tomato – dominant traits, and yellow color and pear-shape – recessive?
2. Prove that if such a crossing is manifested the law of independent distribution of genes.

Solving problems on monohybrid interbreeding with full dominance usually does not cause difficulties.
So we will focus only on the example of the inheritance of a particular trait with incomplete dominance.